A376085 a(0..5) = 1 and a(n) = 1 - a(n-1) - a(n-2) + a(n-1)*a(n-2)*a(n-3)/a(n-4) + a(n-2)*a(n-3)*a(n-4)/a(n-5) + a(n-3)*a(n-4)*a(n-5)/a(n-6), for n > 5.

1, 1, 1, 1, 1, 1, 2, 2, 4, 17, 68, 2312, 668169, 6179226912, 140378107463180352, 250687119058419133437352005889, 325446213917387462112884613611747886778483963398144, 1853431255195849256571682148793108515162996950284389365029788837512893363822697947303936

Offset: 0

Author

Thomas Scheuerle, Sep 09 2024

Keywords

• nonn

Comments

Will this recurrence result into integer values for all n? Yes, because with the start condition a(0..5) = [x, x, x^2, y, y*x^2, y*x^3], we will obtain a sequence of polynomials:

a(6) = (y^3 + y^2 - y)*x^3 + 1.

a(7) = (y^4 + y^3 - y^2)*x^8 + y*x^5.

a(8) = (y^7 + 2*y^6 - y^5 - 2*y^4 + y^3)*x^12 + (2*y^4 + 2*y^3 - 2*y^2)*x^9 + y*x^6 + (-y^2 + y)*x^3.

a(9) = (y^13 + 4*y^12 + 2*y^11 - 8*y^10 - 5*y^9 + 8*y^8 + 2*y^7 - 4*y^6 + y^5)*x^20 + (4*y^10 + 12*y^9 - 20*y^7 + 12*y^5 - 4*y^4)*x^17 + (6*y^7 + 12*y^6 - 6*y^5 - 12*y^4 + 6*y^3)*x^14 + (-y^8 - y^7 + 3*y^6 + y^5 + y^4 + 5*y^3 - 4*y^2)*x^11 + (-2*y^5 + 4*y^3 - 2*y^2 + y)*x^8 + (-y^2 + y)*x^5 + (y^2 - y)*x^3 + 1.

...

For this polynomials a(3*n) divides a(3*n+1).

With the start condition [x, x, x^2, y, y*x^2, y^2*x^3], a(3*n) divides a(3*n+2) too. These polynomials are also more elegant:

a(6) = y^4*x^3 + 1.

a(7) = y^6*x^8 + y^2*x^5.

a(8) = y^11*x^12 + 2*y^7*x^9 + y^3*x^6.

a(9) = y^19*x^20 + 4*y^15*x^17 + 6*y^11*x^14 + 4*y^7*x^11 + y^3*x^8 + 1.

...

Crossrefs

Cf. A051786.

Programs

PARI

a(n) = if( n<0, n = 6-n); if( n<6, 1, 1-a(n-1)-a(n-2)+a(n-1)*a(n-2)*a(n-3)/a(n-4)+a(n-2)*a(n-3)*a(n-4)/a(n-5)+a(n-3)*a(n-4)*a(n-5)/a(n-6))

Formula

a(n) = 1 - a(n-1) - a(n-2) + (a(n-5)^2*a(n-4)^2*a(n-3) + a(n-6)*a(n-4)^2*a(n-3)*a(n-2) + a(n-6)*a(n-5)*a(n-3)*a(n-2)*a(n-1))/(a(n-6)*a(n-5)*a(n-4)).

a(3*n) divides a(3*n+1) and a(3*n+2) too.

a(3*n-1)*a(3*n) divides a(3*n+1) and a(3*n+2).

if the prime p divides a(3*n+1) or a(3*n+2), then it will also divide a(3*n-1)*a(3*n), new prime factors appear the first time in a(3*n) only.